The Quantum Mechanics of Lattice Vibrations

• We’ve seen that the physics of lattice vibrations in a crystalline solid reduces to a CLASSICAL normal mode problem.
The goal of the entire discussion has been to find
the normal mode vibrational frequencies of the solid.
• In the harmonic approximation, this is achieved by first writing the solid’s vibrational energy as a system of coupled simple harmonic oscillators & then finding the classical normal mode frequencies & ion displacements for that system.
• Given the results of these classical normal mode calculations for the vibrating solid, in order to treat some properties of the solid, it is necessary to QUANTIZE these normal modes.

• These quantized normal modes of vibration are called
• PHONONS are massless quantum mechanical particles which have no classical analogue .
They behave like particles in momentum space or k space .
• Phonons are one example of many like this in many areas of physics. Such quantum mechanical particles are often called
“Quasiparticles”
Examples of other Quasiparticles:
Photons: Quantized Normal Modes of electromagnetic waves.
Rotons: Quantized Normal Modes of molecular rotational excitations.
Magnons: Quantized Normal Modes of magnetic excitations in magnetic solids
Excitons: Quantized Normal Modes of electron-hole pairs
Polaritons: Quantized Normal Modes of electric polarization excitations in solids
+ Many Others!!!

PHONONS
• Quantized normal modes of lattice vibrations. The energies & momenta of phonons are quantized
E
 h
 s phonon  p phonon
 h

λ
Phonon wavelength: phonon
≈ a
0
≈ 10 -10 m

PHONONS
• Quantized normal modes of lattice vibrations. The energies & momenta of phonons are quantized
E
 h
 s phonon  p phonon
 h

λ
Phonon wavelength: phonon
≈ a
0
≈ 10 -10 m
PHOTONS
• Quantized normal modes of electromagnetic waves.
The energies & momenta of photons are quantized
E photon
 hc
 p photon
 h

Photon wavelength:
λ
(visible) photon
≈ 10 -6 m

Quantum Mechanical Simple Harmonic Oscillator
• Quantum mechanical results for a simple harmonic oscillator with classical frequency ω : The energy is quantized:
 n



1
2 

 
 n = 0,1,2,3 ,..









Often, we consider
ε n as being constructed by adding n excitation quanta of energy to the ground state.

1
2


Ground state energy of the oscillator.
If the system makes a transition from a lower energy level to a higher energy level, it is always true that the change in energy is an integer multiple of 

Phonon absorption or emission.
ΔE = (n – n΄)

 n & n ΄ = integers
In complicated processes, such as phonons interacting with electrons or photons, it is known that phonons are not conserved.
That is, they can be created and destroyed during such interactions.

As we’ve been discussing in detail, the atoms in a crystal vibrate about their equilibrium positions.
This motion produces vibrational waves.
In a solid, the energy associated with these vibrations is called the Thermal Energy

• A knowledge of the
is fundamental to obtaining an understanding many of the basic properties of solids. A relevant
is how do we calculate this
?
• Also, we would like to know how much thermal energy is available to scatter a conduction electron in a metal or semiconductor. This is important; this scattering contributes to electrical resistance in the material.

• A knowledge of how the thermal energy changes with temperature gives an understanding of the heat energy which is necessary to raise the temperature of the material.
• An important, measureable property of a solid is it’s

The thermal energy is the dominant contribution to the heat capacity in most solids. In non-magnetic insulators, it is the only contribution. Some other contributions are:
Conduction Electrons in metals & semiconductors .
Magnetic ordering in magnetic materials.
A calculation of the vibrational contribution to the thermal energy & heat capacity of a solid has 2 parts:
1.
Evaluation of the contribution of a single vibrational mode.
2.
Summation over the frequency distribution of the modes.

p

•
First, calculate the thermal energy of one mode , then, sum over modes to find the thermal energy due to all modes. From the Canonical ensemble of Statistical
Mechanics , the Average Energy of a harmonic oscillator & hence of a lattice mode of angular frequency ω at temperature T is :
Quantized Energy of a single simple harmonic oscillator:

_
 n
  n
P n
 n

1


2




In the Canonical
Ensemble , the probability of the oscillator being in energy level n at temperature T is proportional to: exp(
  n
/ k T
B
)


_
  n
P n
 n z

_

 n



0
 n
 n



0
1
2
 exp
 exp


 n




 n


1
2 

1
2
/
 k T
B



/ k T
B

 n



0
1
2
 exp[ ( ) ] z
 e
 
/ 2  e

3

/ 2  e

5

/ 2 
.....
z
 e
 
/ 2
(1
 e
 
/  e

2

/ 
.....
z
 e
 
/ 2
(1
 e
 
/
)

1
According to the Binomial expansion for x«1 where x
  
/
(*)

Eqn (*) can be written

_

_

_

_

_
 k T
B
2
 k T
B
2
 k T
B
2
 k T
B
2
 k T
B
2
1
 z
 k T
B
2


T


T ln

 e
 
/ 2 k T
1
 e
 
/

 
T
 ln e
 
/ 2




 e






T




2 k T
B





T

/
 e






2 k
B
4

2 2 k T

 k
B k T
2
B 
1
 e
 e


/

/



 

1
2

/

 
 



1

 e
 e


/

/
Finally, the result is 
_

1
2
  e

/


1


 x
 x
' x


_

1
2
  e

/


1
This is the Mean Phonon Energy.
The first term in the above equation is the zero-point energy . As mentioned before even at 0ºK atoms vibrate in the crystal and have zero-point energy. This is the minimum energy of the system.
The average number of phonons is given by the Bose-Einstein distribution as
(number of phonons) x (energy of phonon) = (second term in )

_ n (

)
 e


1 k
B
T

1
The second term in the mean energy is the phonon contribution to the thermal energy.

1
2

 k
B
T

Mean energy of a harmonic oscillator as a function of T
T
Low Temperature Limit


 k
B
T

_

1
2

  e



 k
B
T

1
Since exponential term gets bigger

_

1
2

 Zero Point Energy

 k
B
T k T
B
1
2



T
High Temperature Limit

<< k T
B is independent of frequency of oscillation.
This is the classical limit because the energy steps are now small compared with the energy of the harmonic oscillator.
So that
 is the thermal energy of the classical 1D harmonic oscillator.
Mean energy of a harmonic oscillator as a function of T

_ e x 
1
 x
 x
2

..........
2 !
e



1
2 k
B
T

_



1

 
1
2


1 k

B
T



 k
B
T
  k T
B

1
_
  k T
B

Heat Capacity C
• Heat capacity C can be found by differentiating the average phonon energy

_

1
2

  e



 k
B
T

1
C v
 d
 dT

 
  k T
B k
 
B
2 e



1
 e
2

C v
 k
B  k T
B

2
2
 e

 e

1

2
Let
  k

C v k
 2

T
 e

T e

1

2

C v
 2

T
 e

T e

1

2 where
  k
 k
B
C v
 k
B
Area =
T
2

Specific heat in this approximation vanishes exponentially at low T and tends to classical value at high temperatures.
These features are common to all quantum systems; the energy tends to the zero-point-energy at low T and to the classical value of Boltzmann constant at high T.

3 R
C v
Specific heat at constant volume depends on temperature as shown in figure below. At high temperatures the value of C v is close to 3R, where R is the universal gas constant. Since R is approximately 2 cal/K-mole, at high temperatures C v is app. 6 cal/K-mole.
This range usually includes RT.
From the figure it is seen that C v equal to 3R at high temperatures is regardless of the substance. This fact is known as Dulong-Petit law. This law states that specific heat of a given number of atoms of any solid is independent of temperature and is the same for all materials !

Points:
Experiment
Curve:
Einstein Model
Prediction
v

The solid is one in which each atom is bound to its side by a harmonic force. When the solid is heated, the atoms vibrate around their sites like a set of harmonic oscillators. The average energy for a 1D oscillator is kT. Therefore, the averaga energy per atom, regarded as a 3D oscillator, is 3kT, and consequently the energy per mole is

Nk T
B

3 RT where N is Avagadro’s number, k
B is Boltzmann constant and R is the gas constant. The differentiation wrt temperature gives;
C v
 d

C v

3 R
   23
3 6.02 10 ( atoms mole
  
23
J K dT
Cv

24.9
J
( K
 mole )
;1 J

0.2388
Cal

Cv 6
Cal
( K
 mole )

• The theory of C v
(T) proposed by Einstein is the first quantum theory of solids. He made the simplifying assumption that all 3N vibrational modes of a 3D solid of N atoms had the same frequency, so that the whole solid has a heat capacity 3N times
C v
 k
B



T


2

T
 e

T e

1

2

• In this model, the atoms are treated as independent oscillators, but the energy of the oscillators are taken quantum mechanically.
• This refers to an isolated oscillator, but the atomic oscillators in a solid are not isolated.They are continually exchanging their energy with their surrounding atoms.
• Even this crude model gave the correct limit at high temperatures, a heat capacity of the Dulong-Petit law where
R is universal gas constant.
Nk
R
B


At high temperatures, all crystalline solids have a specific heat of 6 cal/K per mole ; they require 6 calories per mole to raise their temperature 1 K.
This agreement between observation and classical theory break down if the temperature is low. Experiments show that at room temperature and below the specific heat of crystalline solids is strongly temperature dependent.
C v cal
6
Kmol
 k
B
C 3 R v

In each of these materials
(Pb,Al, Si,and Diamond) specific heat approaches constant value asymptotically at high T. But at low T’s, the specific heat decreases towards zero which is in a
T complete contradiction with the above classical result.

• The Einstein model also gives correctly a specific heat tending to zero at absolute zero , but the temperature dependence near T= 0 did not agree with experiment .
• More accurately taking into account the actual distribution of vibration frequencies in a solid this discrepancy can be accounted for using a modelt due to Peter Debye.

Thermal Energy & Heat Capacity
Debye Model
Density of States
From Quantum Mechanics, if a particle is constrained; the energy of particle can only have discrete energy values. It cannot increase infinitely from one value to another. It has to go up in steps .

• These steps can be so small depending on the system that the energy can be considered as continuous.
• This is the case of classical mechanics.
• But on atomic scale the energy can only jump by a discrete amount from one value to another.
Definite energy levels Steps get small Energy is continuous

• In some cases, each particular energy level can be associated with more than one different state (or wavefunction )
• This energy level is said to be degenerate.
• The density of states is the number of discrete states per unit energy interval , and so is .

There are two sets of waves for solution;
• Running waves
• Standing waves
Running waves:
0 k

4


2

2

4

6

L L L L L
These allowed k wavenumbers corresponds to the running waves; all positive and negative values of k are allowed . By means of periodic boundary condition
L

Na
 p
 
Na

2 p k
 k
2

Na p k
2

L p an integer
Length of the 1D chain
These allowed wavenumbers are uniformly distibuted in k at a
  
.
running waves

R

L
2
 dk

Standing waves: 5

L
4

L
3

L
0

2

3

L L L k
6

L
2

L
7

L
0

L
In some cases it is more suitable to use standing waves,i.e. chain with fixed ends. Therefore we will have an integral number of half wavelengths in the chain;
L
 n

; k
2

2

 k
2
 n
2 L n

L
These are the allowed wavenumbers for standing waves; only positive values are allowed.
k

2

L p for running waves k


L p for standing waves

These allowed k’s are uniformly distributed between k and k+dk at a density of

S

S
( )

L
 dk

R

L
2
 dk
DOS of standing wave
DOS of running wave
•The density of standing wave states is twice that of the running waves .
•However in the case of standing waves only positive values are allowed
•Then the total number of states for both running and standing waves will be the same in a range dk of the magnitude k
•The standing waves have the same dispersion relation as running waves , and for a chain containing N atoms there are exactly N distinct states with k values in the range 0 to

/ a
.

The density of states per unit frequency range g(

):
• The number of modes with frequencies  and

+d
 will be g(

)d

.
• g( 
) can be written in terms of

S
( k ) and

R
( k ).
dR modes with frequency from
 to

+ d
 corresponds dn modes with wavenumber from k to k + dk

dn
 
R
( )
 g
  ;
Choose standing waves to obtain g
  
S g dn
 
S
( )
 g
 
Let’s remember dispersion relation for 1D monoatomic lattice

2 
4 K sin
2 m ka
2 d


2 dk 2 a K ka cos m 2
 
K ka
2 sin m 2 g
 
S k a
1
K ka cos m 2

g
  
S k
1 m a K

1 cos ka / 2
 sin
2 x
 cos
2 x
  cos x

1 sin
2 x cos
 ka

 2 

1 sin
2
 ka

 2  g
  
S k
1 m a K
1
2
 ka

 2 
4
4
Multibly and divide g
  
S k
1 2 a 4 K

4 K m m sin
2
 ka

 2  g
 
L

2 a 
2
1 max
True density of states
 
2
Let’s remember:

S
( )

L
 dk
L

Na
 ka


2 

2  max
4 K m
4 K sin m
2
 2 

g g

2 N



2
 max
 2
 
1/ 2
N m

K
True density of states by means of above equation

  max
2
K m

K m constant density of states
 
K m

Constant density of states can be obtained by ignoring the dispersion of sound at wavelengths comparable to atomic spacing.

The energy of lattice vibrations will then be found by integrating the energy of single oscillator over the distribution of vibration frequencies. Thus
  

0
1
2
  e


/ kT 
1
 g
 
Mean energy of a harmonic oscillator
2 N



2
 max
 2
 
1/ 2 for 1D running waves.
It should be better to find 3D DOS in order to compare the results with experiment.

3D DOS
• Let’s do it first for 2D
• Then for 3D.
• Consider a crystal in the shape of 2D box with crystal lengths of L.
k y y
L
-
+ -
+
+ -
0 x
L
Standing wave pattern for a 2D box

L

L
Configuration in k -space k x

•Let’s calculate the number of modes within a range of wavevector k.
•Standing waves are choosen but running waves will lead same expressions.
•Standing waves will be of the form
U

U
0 sin sin
• Assuming the boundary conditions of
•Vibration amplitude should vanish at edges of x

0; y

0; x

L y

L positive integer
Choosing p
 k
 x
L
; k y
 q

L

y k y
L
0
-
+
+ -
L
-
+ x
Standing wave pattern for a
2D box

L

L
Configuration in k -space k x
 positive quadrant of k -space.
•These values will so be distributed uniformly with a density of
  
2
• This result can be extended to 3D.

L k x
L k z dk k
L
Octant of the crystal: k x
,k y
,k z
(all have positive values)
The number of standing waves;
 s
 
3 k d k

L
3

3 d k

V

3
3 d k k y
L /

 s
 
3
 s
 
3 k d k


V

3
Vk
2
2

2
1

8
1

8
4

4

2 k dk dk

S

Vk
2
2

2

•  
Vk
2
2 is a new density of states defined as the number of
2
 states per unit magnitude of in 3D.This eqn can be obtained by using running waves as well.
• 
(frequency) space can be related to k-space: g
  g
      dk d

Let’s find C at low and high temperature by means of using the expression of g

High and Low Temperature Limits
 
•
3 Nk T
B
Each of the 3N lattice modes of a crystal C
 d

C

3 Nk
B containing N atoms dT
This result is true only if k


B
At low T’s only lattice modes having low frequencies can be excited from their ground states;

Low frequency long
 sound waves
  v k s v s

 k
0
 a k

v s

 k
 k


1
 dk


1 v d v s s and g

2
Vk dk
2

2 d
 g

V


 v s
2
2


2

2
1 v s at low T v depends on the direction and there are two transverse, one s longitudinal acoustic branch: g

V

2
2

1
2 3 v s
 g

V

2
2

2


1

2
3 v v
L T
3


Velocities of sound in longitudinal and transverse direction

  

0
1
2
  e


/ kT 
1
 g
 
Zero point energy =
 z
  

0
1
2
  e


/ kT 

V
1 2


2
2


1

2 v
3
L v
3
T

 d

 



 z

V
2

2


1

2 v 3 v
L
3
T






 

0
 e


3
/ kT 
1
 d



  z

V
2

2


1

2 v
3
L v
3
T


 
4 
4
3

0
 e
15


/ kT


0
3 e

1

3

/ kT d


1
 d
  

0
 
4
3
  3 e x 
1 x
3 k T
B e x x 3

1 dx
0

 dx
 4
15
C v
 d
 dT

2
15

2
V k
B


1

2
3 v v
L T
3
 k T
B
 3
   x

 k T
B
  k T
B x d
  k T
B dx at low temperature

How good is the Debye approximation at low T?
C v
 d
 dT

2
15
V

2 k
B

 v
1

2
3
L v
3
T k T
B
 3
The lattice heat capacity of solids temperatures; this is referred to
T illustrates the excellent agreement of this prediction with experiment for a non-magnetic insulator. The heat capacity vanishes more slowly than the exponential behaviour of a single harmonic oscillator because the vibration spectrum extends down to zero frequency.

for 3D, so it must be calculated numerically.
Debye obtained a good approximation to the resulting heat capacity by neglecting the dispersion of the acoustic waves, i.e. assuming s k for arbitrary wavenumber. In a one dimensional crystal this is density of states figure rather than full curve. Debye’s approximation gives the correct answer in either the high and low temperature limits, and the language associated with it is still widely used today.

The Debye approximation has two main steps:
1. Approximate the dispersion relation of any branch by a linear extrapolation of the small k behaviour:
Einstein approximation to the dispersion
Debye approximation to the dispersion
  vk

Debye cut-off frequency

D
2. Ensure the correct number of modes by imposing a cutoff frequency , above which there are no modes. The
D cut-off freqency is chosen to make the total number of
D modes in a crystal having N atoms, we choose so that

D 
0 g
  
3 N g

V

2

2
2
(
1

2
3 v v
L
3
T
) V
2

2
(
1

2
3 v v
L
3
T
)

D

0

3 N
V
6

2
(
1

2
3 v v
L
3
T
)

3
D

3 N
V
2

2
(
1
 v
3 v
L
2
3
T
)

3

N
3
D
3

9 N

3
D g

9 N

3
D

2 g (

) /

2

The lattice thermal energy is becomes
E
 

0
(
1
2
  e

/


1 g
 
E

9 N

D
3

D 
0
(
1
2
  e

/


1
) d

9 N

D
3
 

 
D
0

2

3 d
 

D 
0 e

/

3

1 d

 

 and,

9
E N
8

D

9 N

D
3

D 
0 e

/ 
1
First term is the estimate of the zero point energy, and all T dependence is in the second term. The heat capacity is obtained by differentiating above eqn wrt temperature.

The heat capacity is

9
E N
8

D

9 N

D
3

0

D e

/
C
 dE dT

1
C
D
 dE dT

9

N
3
D

D 
0
2

4 k T
B
2

/ e
 e

/ 
1
2
 d

Let’s convert this complicated integral into an expression for the specific heat changing variables to x

 k T
B d
 dx
 kT   kT x and define the Debye temperature
 
D

D k
B

The Debye prediction for the lattice specific heat
C
D
 dE dT

9

N
3
D k T k T
B 

B


4
 

 k T
B
2
2




D 
0
/ T x 4 x e
 e x 
1

2 dx
C
D

9 Nk
B


T

D


3 
D 
0
/ T x 4 x e
 e x 
1

2 dx where
 
D

D k
B

C
D
High temperature T

D
T x is always small e x 1 x x
2 x
3
    
2!
3!
 e x
4 x e
 x
1

2

 x
4
(1
 x
1
 
1

)
2
 x
4
(1 x
2
 x )
 x
2
 
D
C
D

9 Nk
B

T
 3


D


D 
0
/ T
2 x dx

3 Nk
B

C
D
Low temperature T

D
For low temperature the upper limit of the integral is infinite; the integral is then a known integral of .

4
4 /15
T
 
D
C
D

9 Nk
B

T
 3


D



D
0
/ T x 4 x e
 e x 
1

2 dx
T
C
D

12 Nk
B
5

4



T


D

 3

Lattice heat capacity due to Debye interpolation scheme
C
D

9 Nk
Figure shows the heat capacity
B


T
 3



D


 D
0
/ T between the two limits of high and low T as predicted by the Debye interpolation formula .
1 x
 e x

1

2 dx
C
3 Nk T
B
Because it is exact in both high and low T limits the Debye formula gives quite a good
Lattice heat capacity of a solid as predicted by the Debye interpolation scheme representation of the heat capacity of most solids, even though the actual phonon-density of T /

D 1 states curve may differ appreciably from the
Debye assumption.
Debye frequency and Debye temperature scale with the velocity of sound in the solid.
So solids with low densities and large elastic moduli have high . Values of for maximum phonon energy in a solid.
Solid

D
Ar
93
Na
158
Cs
38
Fe
457
Cu
343
Pb
105
C KCl
2230 235

v
Points:
Experiment
Curve:
Einstein Model
Prediction

Vibrational Density of States (Aluminum)
Solid Curve:
From X-Ray
Experiment
Dashed Curve:
Debye
Approximation

Debye Density of States

Solid Curve:
Debye
Approximation
Dashed Curve:
Einstein Model
v